Nice Inequality Problem from APMO 2024 P3

投稿日:2024/6/30最終更新日:2025/5/18

Very amazing problem

Problem Statement

Let $n$ be a positive integer and $a_1,a_2,\ldots,a_n$ be positive real numbers. Prove that

$$\sum_{i=1}^{n} \frac{1}{2^i} \left( \frac{2}{1+a_i} \right)^{2^i} \geq \frac{2}{1+a_1a_2\ldots a_n}-\frac{1}{2^n}.$$

Solution

Note that

$$\frac{1}{2^n}+\sum_{i=1}^n \frac{1}{2^i}\left(\frac{2}{1+a_i}\right)^{2^i} = \frac{1}{2^n}+\frac{1}{2^n}\left(\frac{2}{1+a_n}\right)^{2^n}+\sum_{i=1}^{n-1} \frac{1}{2^i}\left(\frac{2}{1+a_i}\right)^{2^i},$$

and by AM-GM inequality,

$$\frac{1}{2^n}+\frac{1}{2^n}\left(\frac{2}{1+a_n}\right)^{2^n}\geq \frac{1}{2^{n-1}}\left(\frac{2}{1+a_n}\right)^{2^{n-1}}.$$

Thus we have

$$\frac{1}{2^n}+\sum_{i=1}^n \frac{1}{2^i}\left(\frac{2}{1+a_i}\right)^{2^i}\geq \frac{1}{2^{n-1}}\left(\frac{2}{1+a_n}\right)^{2^{n-1}}+\sum_{i=1}^{n-1} \frac{1}{2^i}\left(\frac{2}{1+a_i}\right)^{2^i}.$$

Claim

For $x, y\in\mathbb{R}_{0}$ and $m\in\mathbb{Z}_{0},$

$$\cfrac{\left(\frac{2}{1+x}\right)^{2^m}+\left(\frac{2}{1+y}\right)^{2^m}}{2} \ge \left(\frac{2}{1+xy}\right)^{2^{m-1}}.$$

Proof. We proceed by induction. The base case $m=1$ requires us to show that

$$\cfrac{\left(\frac{2}{1+x}\right)^{2}+\left(\frac{2}{1+y}\right)^{2}}{2} \ge \frac{2}{1+xy}.$$

Upon expanding and simplifying it becomes

$$x^3y+xy^3+1\geq x^2y^2+2xy\implies x^2+y^2+\frac{1}{xy}-xy-2\geq 0\implies (x-y)^2+\left(\sqrt{xy}-\frac{1}{\sqrt{xy}}\right)^2\geq 0,$$

which is clearly true.

Suppose the statement is indeed true for some $k\in\mathbb{Z}_{0}.$ Observe that by RMS-AM inequality, we have

$$\cfrac{\left(\frac{2}{1+x}\right)^{2^{k+1}}+\left(\frac{2}{1+y}\right)^{2^{k+1}}}{2}\geq \left(\cfrac{\left(\frac{2}{1+x}\right)^{2^{k}}+\left(\frac{2}{1+y}\right)^{2^{k}}}{2}\right)^2\geq \left(\left(\frac{2}{1+xy}\right)^{2^{k-1}}\right)^2=\left(\frac{2}{1+xy}\right)^{2^k},$$

and so it holds for $k+1$ as well, concluding our proof. $\blacksquare$

Hence, we get that

$$\frac{1}{2^{n-1}}\left(\frac{2}{1+a_n}\right)^{2^{n-1}}+\frac{1}{2^{n-1}}\left(\frac{2}{1+a_{n-1}}\right)^{2^{n-1}}+\sum_{i=1}^{n-2} \frac{1}{2^i}\left(\frac{2}{1+a_i}\right)^{2^i}\geq \frac{1}{2^{n-2}}\left(\frac{2}{1+a_{n-1}a_n}\right)^{2^{n-2}}+\sum_{i=1}^{n-2} \frac{1}{2^i}\left(\frac{2}{1+a_i}\right)^{2^i},$$

and we proceed similarly to establish that

$$\frac{1}{2^n}+\sum_{i=1}^n \frac{1}{2^i}\left(\frac{2}{1+a_i}\right)^{2^i}\geq \frac{2}{1+a_1a_2\ldots a_{n-1}a_n} \implies \sum_{i=1}^{n} \frac{1}{2^i}\left(\frac{2}{1+a_i}\right)^{2^i} \geq \frac{2}{1+a_1a_2\dots a_{n-1}a_n}-\frac{1}{2^n},$$

as desired. $\blacksquare$

Comment Upon initially addressing this problem, I considered applying Tchebycheff's inequality, expressed as

$$\frac{1+a_1a_2\dots a_{n-1}a_n}{2} \geq \frac{(1+a_1)(1+a_2)\dots (1+a_{n-1})(1+a_n)}{2^n}.$$

However, upon further examination, it appears that this inequality does not consistently hold. For instance, when $n=2$, the expression simplifies to $(a_1-1)(a_2-1) \geq 0$, which is not universally true.

Remark There is an easy way to do it. We only need to prove for $\forall 1\le i\le n.$